What is the function that satisfies $\int_0^x f(t) dt=constant$ $$\int_0^x f(t) dt=constant$$
What is the function that satisfies this condition ?
Thank you!
 A: Differentiating both sides of $$\int_0^x f(t) dt=constant$$ using Fundamental Theorem Of Calculus, as mentioned by Omnomnomnom in his comment,( or Leibnitz Rule as I had earlier written),
we get
$$f(x)=0$$
A: Let's denote the continuous function:
$$G(x)=\int_0^x f(t) dt$$
Then by the Fundamental Theorem of Calculus,
$$\frac{d}{dx}G(x)=f(x)$$
When $G(x)=C$, $$\frac{d}{dx}G(x)=0=f(x)$$
A: Because OP has never said $f(t)$ has to be a continuous function, there are more possible functions than a constant zero. For example, if we define $$f(t)=1\text{ for $t\in\Bbb Z$}, 0\text{ otherwise}$$
We get a nonconstant function integral of which is $0$ on every integral. More generally, any function which is non-zero on a discrete set of points will work. 
But these aren't all functions: take, for example, Thomae's function. It is only nonzero on a set of measure zero, and is Riemann-integrable, hence its integral over any interval is again zero.
If we limit ourselves to Riemann-integrable functions, I believe this is the necessary and sufficient condition:

$f(t)=0$ for every $t$ at which $f$ is continuous.

Since Riemann-integrable function must be continuous outside a set of measure zero, this condition is enough to conclude $f$ has integral $0$ on every interval. Conversely, if $f(t_0)>0$ at a point of continuity $t_0$ ($f(t_0)<0$ is treated similarly), then on some open interval we have $f(t)>\frac{f(t_0)}{2}>0$, so on this interval the integral will increase, so it won't be constant.
If we allow more general integrals (Lebesgue measurable) then I believe an equivalent condition would be that $f(t)$ is non-zero only on a set of measure zero, but I don't know enough about Lebesgue-integrability to say that with certainty.
A: Clearly the constant is zero,Because:
$$ constant =   \int_0^0 f(x) dx = 0$$
Can you conclude now?
